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Prove $x^2 + 2y^2 \neq 805, x,y\in\mathbb{Z}$.

We did this in class and, for the life of me, I cannot remember how to finish the problem.

It starts out by taking all of the values to be $\mod5$.

So, $[x^2]_5 + 2[y^2]_5 \neq [805]_5 \neq [0]_5$.

Naturally (heh), the possible values of $[x^2]_5$ are $[0]_5$, $[1]_5$, or $[4]_5$ and it follows for $2[y^2]_5$ the values are $[0]_5$, $[2]_5$, or $[3]_5$.

Thus, all combinations of $[x^2]_5\neq[0]_5\neq2[y^2]_5$, the equation equals everthing BUT $[0]_5$. Which is good.

However, then we reach a point where we must ask ourselves why it is impossible in the equation for $[x]_5$ and $[y]_5$ to be congruent to $[0]_5$. This is the part I don't remember how to do: How do I prove that $[0]_5 + [0]_5 \neq [0]_5$ is a contradiction/does not apply for $x^2 + 2y^2 = 805$?

I do remember it being a rather simple contradiction, but I cannot think of it.

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I don't know much on the topic, but sense there are only a finite number of integers such that x^2+2y^2<805, wouldn't it just be faster to check all the possible solutions to show that none of them satisfy the equation, say by cycling fixed integer values of y, and checking that their isn't an x that satisfies the equation, I don't think it would require more then the checking of roughly 10-20 values of y, with coresponding x's. –  Ethan Nov 29 '12 at 8:43
    
It's a proof based math course that I have a test in tomorrow. I was reviewing and came across this problem and got stuck. I knew I could have done an exhaustive search at this point, but that's not very good for taking a test. –  Logan Nov 29 '12 at 8:52

1 Answer 1

up vote 11 down vote accepted

Hint: If $x^2$ and $y^2$ are divisible by $5$, then so are $x$ and $y$, and $x^2$ and $y^2$ are then divisible by $25$.

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Of course. Thank you. I would vote up if I could. :) –  Logan Nov 29 '12 at 8:46

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